Correct Answer: $462-308\sqrt{2}$

Solution :

Let the radius of the circle = $r$ cm
The radius of the quadrant = QM = QR = QP = 7 cm
From the figure,
$OS=OT=OM=QT=r$ cm
⇒ $OQ=(7-r)$ cm
In $\triangle QTO$,
$OQ^2=OT^2+QT^2$
⇒ $(7-r)^2=r^2+r^2$
⇒ $(7-r)^2=2r^2$
⇒ $7-r=r\sqrt2$
⇒ $r=\frac{7}{\sqrt2+1}=7(\sqrt2-1)$
The area of the circle = $\pi r^2$
⇒ The area of the circle = $\pi [7(\sqrt2-1)]^2$
⇒ The area of the circle = $\frac{22}{7} [7(\sqrt2-1)]^2$
⇒ The area of the circle = $22×7×(3-2\sqrt2)=462-308\sqrt{2}$ cm ²
Hence, the correct answer is $462-308\sqrt{2}$.